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At 298 k 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water.the ph of resulting solution is ? A 4.9 B 4.6 C 4.3 D 4.2?
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At 298 k 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are d...
Solution:
The given problem is a typical example of a weak acid-strong base buffer solution. The buffer solution is a solution that resists the change in pH when an acid or a base is added to it. The buffer solution is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. In this problem, the weak acid is acetic acid and its conjugate base is acetate ion.
Step 1: Write the balanced chemical equation for the dissociation of ammonium acetate and acetic acid.
NH4C2H3O2 ⇌ NH4+ + C2H3O2-
CH3COOH ⇌ CH3COO- + H+
Step 2: Calculate the initial concentration of acetic acid and acetate ion.
Initial concentration of acetic acid (CH3COOH) = 0.14 mol/L
Initial concentration of acetate ion (C2H3O2-) = 0 mol/L
Step 3: Calculate the concentration of ammonium ion and acetate ion after the dissociation of ammonium acetate.
NH4C2H3O2 ⇌ NH4+ + C2H3O2-
Initial concentration = 0.1 mol/L 0 0
Change -x +x +x
Equilibrium concentration 0.1-x x x
Step 4: Write the equilibrium constant (Ka) expression for the dissociation of acetic acid.
Ka = [CH3COO-][H+]/[CH3COOH]
Step 5: Substitute the equilibrium concentrations of acetic acid, acetate ion, and hydrogen ion into the Ka expression.
Ka = (x)(x)/(0.14-x)
Step 6: Solve for x using the quadratic formula.
Ka = 1.8 × 10^-5
x = 7.5 × 10^-4 mol/L
Step 7: Calculate the pH of the buffer solution.
pH = pKa + log([base]/[acid])
pH = 4.76 + log(0.75/0.05)
pH = 4.9
Therefore, the pH of the resulting solution is option A, 4.9.
The given problem is a typical example of a weak acid-strong base buffer solution. The buffer solution is a solution that resists the change in pH when an acid or a base is added to it. The buffer solution is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. In this problem, the weak acid is acetic acid and its conjugate base is acetate ion.
Step 1: Write the balanced chemical equation for the dissociation of ammonium acetate and acetic acid.
NH4C2H3O2 ⇌ NH4+ + C2H3O2-
CH3COOH ⇌ CH3COO- + H+
Step 2: Calculate the initial concentration of acetic acid and acetate ion.
Initial concentration of acetic acid (CH3COOH) = 0.14 mol/L
Initial concentration of acetate ion (C2H3O2-) = 0 mol/L
Step 3: Calculate the concentration of ammonium ion and acetate ion after the dissociation of ammonium acetate.
NH4C2H3O2 ⇌ NH4+ + C2H3O2-
Initial concentration = 0.1 mol/L 0 0
Change -x +x +x
Equilibrium concentration 0.1-x x x
Step 4: Write the equilibrium constant (Ka) expression for the dissociation of acetic acid.
Ka = [CH3COO-][H+]/[CH3COOH]
Step 5: Substitute the equilibrium concentrations of acetic acid, acetate ion, and hydrogen ion into the Ka expression.
Ka = (x)(x)/(0.14-x)
Step 6: Solve for x using the quadratic formula.
Ka = 1.8 × 10^-5
x = 7.5 × 10^-4 mol/L
Step 7: Calculate the pH of the buffer solution.
pH = pKa + log([base]/[acid])
pH = 4.76 + log(0.75/0.05)
pH = 4.9
Therefore, the pH of the resulting solution is option A, 4.9.
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At 298 k 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water.the ph of resulting solution is ? A 4.9 B 4.6 C 4.3 D 4.2?
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At 298 k 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water.the ph of resulting solution is ? A 4.9 B 4.6 C 4.3 D 4.2? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about At 298 k 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water.the ph of resulting solution is ? A 4.9 B 4.6 C 4.3 D 4.2? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 298 k 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water.the ph of resulting solution is ? A 4.9 B 4.6 C 4.3 D 4.2?.
At 298 k 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water.the ph of resulting solution is ? A 4.9 B 4.6 C 4.3 D 4.2? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about At 298 k 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water.the ph of resulting solution is ? A 4.9 B 4.6 C 4.3 D 4.2? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 298 k 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water.the ph of resulting solution is ? A 4.9 B 4.6 C 4.3 D 4.2?.
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